nth Roots Of Unity  z

n

 1
nth Roots Of Unity  z

n

 1

The solutions of equations of the form, z n  1, are the nth roots of unity
nth Roots Of Unity  z

n

 1

The solutions of equations of the form, z n  1, are the nth roots of unity
When placed...
nth Roots Of Unity  z

n

 1

The solutions of equations of the form, z n  1, are the nth roots of unity
When placed...
nth Roots Of Unity  z

n

 1

The solutions of equations of the form, z n  1, are the nth roots of unity
When placed...
nth Roots Of Unity  z

n

 1

The solutions of equations of the form, z n  1, are the nth roots of unity
When placed...
nth Roots Of Unity  z

n

 1

The solutions of equations of the form, z n  1, are the nth roots of unity
When placed...
nth Roots Of Unity  z

n

 1

The solutions of equations of the form, z n  1, are the nth roots of unity
When placed...
nth Roots Of Unity  z

n

 1

The solutions of equations of the form, z n  1, are the nth roots of unity
When placed...
nth Roots Of Unity  z

n

 1

The solutions of equations of the form, z n  1, are the nth roots of unity
When placed...
b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1

If  is a so...
b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1

If  is a so...
b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1

If  is a so...
b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1

If  is a so...
b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1

If  is a so...
b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are
the other roots.
z5  1

If  is a so...
ii  Prove that 1     2   3   4  0
ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2

3

4

5

sum of the roots ...
ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2

3

4

5

5
OR   1  0
 ...
ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2

3

4

sum of the roots 

5...
ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2

3

4

5

sum of the roots ...
ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2

3

4

5

sum of the roots ...
ii  Prove that 1     2   3   4  0
b
      
a
1   2  3  4  0
2

3

4

5

sum of the roots ...
c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
(i ) Evaluate 1  



2 3
c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
(i ) Evaluate 1  
   



2 3

3
c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
(i ) Evaluate 1  
   

3

  3
 ...
c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1 ...
c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1 ...
c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1 ...
c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1 ...
c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1 ...
c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1 ...
c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to;
1
1
2 3

(ii ) Evaluate
(i ) Evaluate 1 ...
d) Solve z 5  z 4  z 3  z 2  z  1  0
d) Solve z 5  z 4  z 3  z 2  z  1  0

Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
d) Solve z 5  z 4  z 3  z 2  z  1  0

Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
And z 6  1  0 has solu...
d) Solve z 5  z 4  z 3  z 2  z  1  0

Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
And z 6  1  0 has solu...
d) Solve z 5  z 4  z 3  z 2  z  1  0

Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
And z 6  1  0 has solu...
d) Solve z 5  z 4  z 3  z 2  z  1  0

Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
And z 6  1  0 has solu...
e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1
e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1

 ...
e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1

 ...
e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1

 ...
e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1

 ...
e) 1996 HSC
2
2
 i sin
Let   cos
9
9
(i) Show that  k is a solution of z 9  1  0, where k is an integer
z9  1

 ...


2
4 1
(iii) Hence show that cos cos
cos

9
9
9 8


2
4 1
(iii) Hence show that cos cos
cos

9
9
9 8
9
z 1
  z  1 z     z   8  z   2  z   7  z  ...


2
4 1
(iii) Hence show that cos cos
cos

9
9
9 8
9
z 1
  z  1 z     z   8  z   2  z   7  z  ...


2
4 1
(iii) Hence show that cos cos
cos

9
9
9 8
9
z 1
  z  1 z     z   8  z   2  z   7  z  ...
2

i  1   i  1  2cos
9

9

4

i  2cos
9


8
 
i   i   2cos
9
 


i

2

i  1   i  1  2cos
9

9

4

i  2cos
9


8
 
i   i   2cos
9
 

2 
4 
8 

i  1  i ...
2

i  1   i  1  2cos
9

9

4

i  2cos
9


8
 
i   i   2cos
9
 

2 
4 
8 

i  1  i ...
2

i  1   i  1  2cos
9

9

4

i  2cos
9


8
 
i   i   2cos
9
 

2 
4 
8 

i  1  i ...
2

i  1   i  1  2cos
9

9

4

i  2cos
9


8
 
i   i   2cos
9
 

2 
4 
8 

i  1  i ...
2

i  1   i  1  2cos
9

9

4

i  2cos
9


8
 
i   i   2cos
9
 


i


2 
4 
8 

i ...
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X2 t01 11 nth roots of unity (2012)

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X2 t01 11 nth roots of unity (2012)

  1. 1. nth Roots Of Unity  z n  1
  2. 2. nth Roots Of Unity  z n  1 The solutions of equations of the form, z n  1, are the nth roots of unity
  3. 3. nth Roots Of Unity  z n  1 The solutions of equations of the form, z n  1, are the nth roots of unity When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle.
  4. 4. nth Roots Of Unity  z n  1 The solutions of equations of the form, z n  1, are the nth roots of unity When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle. e.g . z 5  1
  5. 5. nth Roots Of Unity  z n  1 The solutions of equations of the form, z n  1, are the nth roots of unity When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle. e.g . z 5  1  2 k  0  z  cis   k  0, 1, 2  5 
  6. 6. nth Roots Of Unity  z n  1 The solutions of equations of the form, z n  1, are the nth roots of unity When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle. e.g . z 5  1  2 k  0  z  cis   k  0, 1, 2  5  2 2 4 4 z  cis 0, cis , cis  , cis , cis  5 5 5 5
  7. 7. nth Roots Of Unity  z n  1 The solutions of equations of the form, z n  1, are the nth roots of unity When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle. e.g . z 5  1  2 k  0  z  cis   k  0, 1, 2  5  2 2 4 4 z  cis 0, cis , cis  , cis , cis  y 5 5 5 5 cis 0 x
  8. 8. nth Roots Of Unity  z n  1 The solutions of equations of the form, z n  1, are the nth roots of unity When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle. e.g . z 5  1  2 k  0  z  cis   k  0, 1, 2  5  2 2 4 4 z  cis 0, cis , cis  , cis , cis  y 2 cis 5 5 5 5 5 cis 0 x 2 cis  5
  9. 9. nth Roots Of Unity  z n  1 The solutions of equations of the form, z n  1, are the nth roots of unity When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle. e.g . z 5  1  2 k  0  z  cis   k  0, 1, 2  5  2 2 4 4 z  cis 0, cis , cis  , cis , cis  y 2 4 cis 5 5 5 5 cis 5 5 cis 0 x 4 cis  5 2 cis  5
  10. 10. nth Roots Of Unity  z n  1 The solutions of equations of the form, z n  1, are the nth roots of unity When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle. e.g . z 5  1  2 k  0  z  cis   k  0, 1, 2  5  2 2 4 4 z  cis 0, cis , cis  , cis , cis  y 2 4 cis 5 5 5 5 cis 5 5 cis 0 x 4 cis  5 2 cis  5
  11. 11. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are the other roots.
  12. 12. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are the other roots. z5  1 If  is a solution then  5  1
  13. 13. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are the other roots. z5  1 If  is a solution then  5  1 5 5     15  5 is a solution 1
  14. 14. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are the other roots. z5  1 If  is a solution then  5  1 5 5     15  5 is a solution 1      2 5 5 2  12 1  2 is a solution
  15. 15. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are the other roots. z5  1 If  is a solution then  5  1 5 5     15  5 is a solution 1      2 5 5 2  12 1  2 is a solution      3 5 5 3  13 1  3 is a solution
  16. 16. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are the other roots. z5  1 If  is a solution then  5  1 5 5     15  5 is a solution 1      2 5 5 2  12 1  2 is a solution      3 5 5 3  13 1  3 is a solution      4 5 5 4  14 1  4 is a solution
  17. 17. b) i  If  is a complex root of z 5  1  0, show that  2 ,  3 ,  4 and  5 are the other roots. z5  1 If  is a solution then  5  1 5 5     15  5 is a solution 1      2 5 5 2  12 1  2 is a solution      3 5 5 3  13 1  3 is a solution      4 5 5 4  14 1  4 is a solution Thus if  is a root then  2 ,  3 ,  4 and  5 are also roots
  18. 18. ii  Prove that 1     2   3   4  0
  19. 19. ii  Prove that 1     2   3   4  0 b        a 1   2  3  4  0 2 3 4 5 sum of the roots   5  1
  20. 20. ii  Prove that 1     2   3   4  0 b        a 1   2  3  4  0 2 3 4 5 5 OR   1  0   11     2   3   4   0 1     2   3   4  0   1 sum of the roots    1  NOTE :   n   1  0     11     2     n1   0   5
  21. 21. ii  Prove that 1     2   3   4  0 b        a 1   2  3  4  0 2 3 4 sum of the roots  5  5 OR   1  0   11     2   3   4   0 1     2   3   4  0   1 OR 1   2  3  4   a  r n  1 r 1 1 5  1  1  1  NOTE :   n   1  0     11     2     n1   0   5 GP: a  1, r   , n  5 0  5 1  0
  22. 22. ii  Prove that 1     2   3   4  0 b        a 1   2  3  4  0 2 3 4 5 sum of the roots   5 OR   1  0   11     2   3   4   0 1     2   3   4  0   1 OR 1   2  3  4  a  r n  1 r 1 1 5  1  1  NOTE :   n   1  0     11     2     n1   0   5 GP: a  1, r   , n  5  1  0  0  1 iii  Find the quadratic equation whose roots are    4 and  2   3 5
  23. 23. ii  Prove that 1     2   3   4  0 b        a 1   2  3  4  0 2 3 4 5 sum of the roots   5 OR   1  0   11     2   3   4   0 1     2   3   4  0   1 OR 1   2  3  4  a  r n  1 r 1 1 5  1  1  NOTE :   n   1  0     11     2     n1   0   5 GP: a  1, r   , n  5  1  0  0  1 iii  Find the quadratic equation whose roots are    4 and  2   3 4 2 3      4  2   3           3  4  6  7  1  3  4    2  1 5
  24. 24. ii  Prove that 1     2   3   4  0 b        a 1   2  3  4  0 2 3 4 5 sum of the roots   5 OR   1  0   11     2   3   4   0 1     2   3   4  0   1 OR 1   2  3  4  a  r n  1 r 1 1 5  1  1  NOTE :   n   1  0     11     2     n1   0   5 GP: a  1, r   , n  5  1  0  0  1 iii  Find the quadratic equation whose roots are    4 and  2   3 4 2 3      4  2   3           3  4  6  7  1  3  4    2  equation is x 2  x  1  0  1 5
  25. 25. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to; (i ) Evaluate 1    2 3
  26. 26. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to; (i ) Evaluate 1        2 3 3
  27. 27. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to; (i ) Evaluate 1       3   3  1  2 3
  28. 28. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to; 1 1 2 3  (ii ) Evaluate (i ) Evaluate 1    1  1 2 3       3  1
  29. 29. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to; 1 1 2 3  (ii ) Evaluate (i ) Evaluate 1    1  1 2 3     1 1  2  3      1  1   2
  30. 30. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to; 1 1 2 3  (ii ) Evaluate (i ) Evaluate 1    1  1 2 3     1 1  2  3      1  1   2 2  2  1
  31. 31. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to; 1 1 2 3  (ii ) Evaluate (i ) Evaluate 1    1  1 2 3     1 1  2  3      1 (iii) Form the cubic equation with roots 1, 1   , 1   2  1   2 2  2  1
  32. 32. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to; 1 1 2 3  (ii ) Evaluate (i ) Evaluate 1    1  1 2 3     1 1  2  3      1 (iii) Form the cubic equation with roots 1, 1   , 1   2  z  1z 2  2     2 z  1     2   3   0  1   2 2  2  1
  33. 33. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to; 1 1 2 3  (ii ) Evaluate (i ) Evaluate 1    1  1 2 3     1 1  2  3      1 (iii) Form the cubic equation with roots 1, 1   , 1   2  z  1z 2  2     2 z  1     2   3   0  z  1z 2  z  1  0  1   2 2  2  1
  34. 34. c) If  is a complex cube root of unity, use the fact that 1  ω  ω 2  0 to; 1 1 2 3  (ii ) Evaluate (i ) Evaluate 1    1  1 2 3     1 1  2  3      1 (iii) Form the cubic equation with roots 1, 1   , 1   2  z  1z 2  2     2 z  1     2   3   0  z  1z 2  z  1  0 z  z  z  z  z 1  0 z3  2z 2  2z 1  0 3 2 2  1   2 2  2  1
  35. 35. d) Solve z 5  z 4  z 3  z 2  z  1  0
  36. 36. d) Solve z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1
  37. 37. d) Solve z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6 
  38. 38. d) Solve z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0
  39. 39. d) Solve z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0  z  1z 5  z 4  z 3  z 2  1 0  z  1 z 6  1  0, z  1
  40. 40. d) Solve z 5  z 4  z 3  z 2  z  1  0 Now z 6  1   z  1z 5  z 4  z 3  z 2  z  1 And z 6  1  0 has solutions;  2 k  z  cis   k  0, 1, 2,3  6  z5  z4  z3  z2 1  0  z  1z 5  z 4  z 3  z 2  1 0  z  1 z 6  1  0, z  1   2 2 z  cis , cis  , cis , cis  , cis 3 3 3 3 1 3 1 3 1 3 1 3 z  i,  i,   i,   i, 1 2 2 2 2 2 2 2 2
  41. 41. e) 1996 HSC 2 2  i sin Let   cos 9 9 (i) Show that  k is a solution of z 9  1  0, where k is an integer
  42. 42. e) 1996 HSC 2 2  i sin Let   cos 9 9 (i) Show that  k is a solution of z 9  1  0, where k is an integer z9  1
  43. 43. e) 1996 HSC 2 2  i sin Let   cos 9 9 (i) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis    9  k 2  z  cis   9   k  0,1,2,3,4,5,6,7,8
  44. 44. e) 1996 HSC 2 2  i sin Let   cos 9 9 (i) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis    9  k 2  z  cis   9   z  k k  0,1,2,3,4,5,6,7,8
  45. 45. e) 1996 HSC 2 2  i sin Let   cos 9 9 (i) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8   9  k 2  z  cis   9   z  k (ii) Prove that    2   3   4   5   6   7   8  1 z9 1  0
  46. 46. e) 1996 HSC 2 2  i sin Let   cos 9 9 (i) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8   9  k 2  z  cis   9   z  k (ii) Prove that    2   3   4   5   6   7   8  1 z9 1  0 1     2   3   4   5   6   7   8  0 sum of roots 
  47. 47. e) 1996 HSC 2 2  i sin Let   cos 9 9 (i) Show that  k is a solution of z 9  1  0, where k is an integer z9  1  2k  z  cis  k  0,1,2,3,4,5,6,7,8   9  k 2  z  cis   9   z  k (ii) Prove that    2   3   4   5   6   7   8  1 z9 1  0 1     2   3   4   5   6   7   8  0 sum of roots     2   3   4   5   6   7   8  1
  48. 48.  2 4 1 (iii) Hence show that cos cos cos  9 9 9 8
  49. 49.  2 4 1 (iii) Hence show that cos cos cos  9 9 9 8 9 z 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5 
  50. 50.  2 4 1 (iii) Hence show that cos cos cos  9 9 9 8 9 z 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5  2 4  2  2  z  1 z  2cos z  1   z  1  z  2cos 9 9    6 8  2   z  2cos z  1 z 2  2cos z  1  9 9    2 4  2  2  z  1 z  2cos z  1   z  1  z  2cos 9 9    8  2  2  z  z  1  z  2cos 9 z  1  
  51. 51.  2 4 1 (iii) Hence show that cos cos cos  9 9 9 8 9 z 1   z  1 z     z   8  z   2  z   7  z   3  z   6  z   4  z   5  2 4  2  2  z  1 z  2cos z  1   z  1  z  2cos 9 9    6 8  2   z  2cos z  1 z 2  2cos z  1  9 9    2 4  2  2  z  1 z  2cos z  1   z  1  z  2cos 9 9    8  2  2  z  z  1  z  2cos 9 z  1   Let z  i 2  i  1   i  1  2cos 9  9 4  i  2cos 9  8   i   i   2cos 9    i 
  52. 52. 2  i  1   i  1  2cos 9  9 4  i  2cos 9  8   i   i   2cos 9    i 
  53. 53. 2  i  1   i  1  2cos 9  9 4  i  2cos 9  8   i   i   2cos 9   2  4  8   i  1  i  i  1  2cos  2cos  2cos  9  9  9   4  i 
  54. 54. 2  i  1   i  1  2cos 9  9 4  i  2cos 9  8   i   i   2cos 9   2  4  8   i  1  i  i  1  2cos  2cos  2cos  9  9  9   2 4 8 cos cos 1  8cos 9 9 9 4  i 
  55. 55. 2  i  1   i  1  2cos 9  9 4  i  2cos 9  8   i   i   2cos 9   2  4  8   i  1  i  i  1  2cos  2cos  2cos  9  9  9   2 4 8 cos cos 1  8cos 9 9 9 2 4   cos 1  8cos   cos  9 9  9 4  i 
  56. 56. 2  i  1   i  1  2cos 9  9 4  i  2cos 9  8   i   i   2cos 9   2  4  8   i  1  i  i  1  2cos  2cos  2cos  9  9  9   2 4 8 cos cos 1  8cos 9 9 9 2 4   cos 1  8cos   cos  9 9  9 4  2 4 1 cos cos cos  9 9 9 8  i 
  57. 57. 2  i  1   i  1  2cos 9  9 4  i  2cos 9  8   i   i   2cos 9    i  2  4  8   i  1  i  i  1  2cos  2cos  2cos  9  9  9   2 4 8 cos cos 1  8cos 9 9 9 2 4   cos 1  8cos   cos  9 9  9 4  2 4 1 cos cos cos  9 9 9 8 Cambridge: Exercise 7C; 1 to 4, 5ac, 6, 7, 8, 9, 11 Patel: Exercise 4I; 1, 3, 5 (need 4b), 7 Patel: Exercise 4J; 1 to 4, 7ac
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